This category collects chemistry questions spanning physical, organic and inorganic topics. Use it to test how well you recall key concepts, balance and interpret reactions, and apply formulas under exam conditions. Each question comes with an answer so you can see where your reasoning went off track and revise that area again.
Co³⁺ with 4 NH₃ and 2 H₂O (neutral) forms [Co(NH₃)4(H₂O)2]^{3+, balanced by 3 Cl^-, giving [Co(NH₃)4(H₂O)2]Cl₃ . This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
[Co(NH₃)5(SCN)]^{2+ has the ambidentate ligand SCN^-, which can bind via S or N, showing linkage isomerism. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
[Au(CN)2]^{- is used in gold electroplating for smooth deposition, unlike the others. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The initial rate of a reaction is 1.6 × 10⁻³mol L^{-1 s^{-1 when [A] = 0.4 M and [B] = 0.1 M. If the rate law is Rate = k[A]², what is k in L mol^{-1 s^{-1 ? These values define the system as per NCERT data. Formula: k = fracRate[A]² = frac1.6 × 10⁻³(0.4)² = frac1.6 × 10⁻³⁰.16 = 0.01 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: A gaseous reaction follows Rate = k p_A² with k = 0.1 atm^{-1 s^{-1 and p_A = 0.6 atm . What is the rate in atm s^{-1 ? These values define the system as per NCERT data. Formula: Rate = k p_A² = 0.1 × (0.6)² = 0.1 × 0.36 = 0.036 .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the molarity of a solution prepared by dissolving 4 g of NaOH in 500 mL of solution? (Molar mass: NaOH = 40 g/mol ) These values define the system as per NCERT data. Formula: Moles of NaOH = 4/40 = 0.1 mol. This is standard NCERT relation. Substitution & Calculation: Volume in liters = 500/1000 = 0.5 L . Molarity = 0.1/0.5 = 0.2 M . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What is the EMF of a ll with Ni(s) | Ni²⁺(0.005 M) || Ag⁺(0.001 M) | Ag(s) at 298 K, given E_{Ni^{2+/Ni⁰ = -0.25 V and E_{Ag^{+/Ag⁰ = 0.80 V ? These values define the system as per NCERT data. Formula: E_{ll⁰ = 0.80 - (-0.25) = 1.05 V, n = 2. This is standard NCERT relation. Substitution & Calculation: E_{ll = E_{ll⁰ - 0.059/2 log frac[Ni^{2+][Ag^{+]² . Q = 0.005/(0.001)² = 5000, log Q = 3.699 . E_{ll = 1.05 - 0.059/2 × 3.699 = 1.05 - 0.109 = 0.941 V approx 0.94 V . Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
CO₂ has 2 double bonds (each with 1 sigma and 1 pi), totaling 2 pi bonds. This follows from NCERT principle where relation explains outcome clearly for students.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The solubility of PbCl₂ in water is 1.0 × 10⁻²M. What is its K_{sp ? These values define the system as per NCERT data. Formula: PbCl₂(s) Pb²+ + 2Cl-, [Pb²+] = S, [Cl-] = 2S. This is standard NCERT relation. Substitution & Calculation: K_{sp = [Pb²+][Cl-]² = S · (2S)² = 4S³ = 4 (1.0 × 10⁻²)³ = 4 × 10⁻⁶. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: Calculate the de Broglie wavelength of an electron moving at 2.05 × __10POW₇__m s^{-1 . ( h = 6.626 × 10⁻³⁴J s, m_e = 9.1 × 10⁻³¹kg ) These values define the system as per NCERT data. Formula: lambda = h/mv = frac6.626 × 10⁻³⁴⁹.1 × 10⁻³¹ × 2.05 × __10POW₇__= 3.55 × 10⁻¹¹m .. This is standard NCERT relation. Substitution & Calculation: Substituting values and simplifying step by step as per NCERT method. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: The heat capacity of a calorimeter is 10 kJ/K. If the temperature rises from 298 K to 300 K during a reaction, what is the heat absorbed by the calorimeter? These values define the system as per NCERT data. Formula: q = C Δ T. This is standard NCERT relation. Substitution & Calculation: C = 10 kJ/K, Δ T = 300 - 298 = 2 K. So, q = 10 × 2 = 20 kJ. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
Given: What volume of 1.5 M HCl is required to react completely with 10 g of CaCO₃? (Molar mass of CaCO₃ = 100 g/mol) These values define the system as per NCERT data. Formula: Moles of CaCO₃ = 10 / 100 = 0.1 mol. This is standard NCERT relation. Substitution & Calculation: Reaction: CaCO₃ + 2HCl → CaCl₂ + CO₂ + H₂O. . Moles of HCl = 0.1 × 2 = 0.2 mol. Volume = 0.2 / 1.5 = 0.1333 L = 133.3 mL ≈ 133 mL. Result: The computed value matches expected outcome and confirms correct choice as per NCERT.
Ref: NCERT Chemistry Textbook for Class XI and XII, Chapter: Some Basic Concepts, Structure of Atom, Periodicity and relevant Chemistry topic, Topic: Mole concept and periodic trends.
